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**ThePerfectHacker** The tangent plane to a sphere is perpendicular to the radius.

Let $\displaystyle (x_0,y_0,z_0)$ be a point on the sphere such that condition (a) is satisfied.

That means, $\displaystyle (x_0-2)^2+(y_0-3)^2+(z_0-3)^2 = 16$, since it is on the sphere.

It also means the radius vector is $\displaystyle (x_0-2)\bold{i}+(y_0-3)\bold{j}+(z_0-3)\bold{k}$. And this is perpendicular to the tangent plane. So it means this vector is the **normal** of the tangent plane.

Now if we want this to be parallel with (a), the xy-plane, then we need the tangent vector to be parallel with the tangent vector of the xy-plane (which is $\displaystyle \bold{k}$). That means $\displaystyle (x_0-2)\bold{i}+(y_0-3)\bold{j}+(z_0-3)\bold{k} = k(\bold{k})$

So, $\displaystyle x_0 = 2, y_0 = 3, z_0 = k+3$.

If you subsitute that into the sphere equation we get,

$\displaystyle (k+3-3)^2 = 16 \implies k = \pm 4$.

So $\displaystyle (2,3,-1)\mbox{ or }(2,3,7)$. At these point the tangent plane is parallel with the xy-plane.

Try the next ones.